MODELING MECHANICAL AND ELECTRICAL
2.3 BASIC MECHANICAL AND ELECTRICAL ELEMENTS
2.3.2 Bending Beams
Abeamis a mechanical element that supports a load perpendicular to its length. The theory discussed in this section applies to beams whose length is much greater than either their width or their thickness. Under these circumstances we will be able to
1 3
cross-sectional area A Young's Modulus Y
L T1 T1
stress state strain state
load
ψ
Figure 2.14 States of stress and strain for a bending beam.
derive a relationship between the applied load and the deflection of the beam and the stored mechanical energy.
The beam geometry that we consider is shown in Figure 2.14. We assume that the length of the beam is aligned with the 1 axis of the Cartesian coordinate system and that any loads are applied in the 3 direction. The material properties of the beam are as- sumed to be homogeneous. Making the kinematic assumptions that there are no dimen- sional changes in the 3 direction and that any section that is plane before deformation remains a plane section after deformation, displacement in the 1 direction is written as
u1=u0+ψx3, (2.121)
whereψis the slope of the plane as shown in Figure 2.14 andu0is the uniform axial displacement. Assuming small strains, the slope of the plane is approximately equal to the first derivative of the displacement in the 3 direction,u3. Assuming that the uniform axial displacement is equal to zero yields
u1=x3du3
d x1. (2.122)
The strain S1is obtained from equation (2.14):
S1= du1
d x1 =x3
d2u3
d2x1. (2.123)
Assuming that each differential element of the material is under uniaxial stress, the strain in the 2 and 3 directions is
S2= −νS1= −νx3
d2u3
d2x1
(2.124) S3= −νS1= −νx3
d2u3
d2x1
,
and the only nonzero stress is given by
T1=x3Yd2u3
d2x1
. (2.125)
The moment caused by the applied load is in equilibrium with the moment produced by the induced stress. The relationship is
M2=
A
x3T1d A, (2.126)
whered Ais the differential area elementd x2d x3. Substituting equation (2.125) into equation (2.126) and taking the terms that are independent of x2andx3 out of the integral, we have
M2=
Yd2u3
d2x1
A
x32d A. (2.127)
We denote the integral as the second area moment of inertia, in this case about the x3axis, and write the relationship as
M2=YI33
d2u3
d2x1. (2.128)
Equation (2.128) is a direct relationship between the moment due to applied loads and the resulting deflection.
Determining the deflection along the length of the beam requires an explicit def- inition of the moment and a set of prescribed boundary conditions. The first case to consider is the case in which the moment is constant along the length. Under this assumption, equation (2.128) can be integrated once to yield
YI33
du3 d x1
=M2x1+c1, (2.129)
wherec1 is a constant of integration. Integrating once more and dividing byYI33
yields an expression for the displacement:
u3(x1)= 1 YI33
1
2M2x12+c1x1+c2 . (2.130) The displacement of the beam in the 3 direction varies as a quadratic function along the length of the beam.
The exact expression for the displacement is a function of the boundary conditions.
Two common boundary conditions areclamped–freeandpinned–pinned. These two
boundary conditions impose the following constraints on the displacement function:
clamped–free: u3(0)= du3 d x1
x1=0=0
(2.131) pinned–pinned: u3(0)=0u3(L)=0.
Substituting the boundary conditions into equations (2.129) and (2.130) yields the displacement functions:
clamped–free: u3(x1)= M2 2YI33x21
(2.132) pinned–pinned: u3(x1)= M2
2YI33
x12−L x1
.
Another loading condition that is common to the problems that we will study later is that of a point load applied along the length of the beam. First consider the case of a clamped–free beam with a point load applied at the free end. The moment induced by this load is
M2(x1)= f(L−x1), (2.133) and the first integral of equation (2.128) is
YI33du3
d x1 = f
L x1−1
2x12 +c1. (2.134)
Integrating equation (2.134) once again yields YI33u3(x1)= f
1
2L x12−1
6x31 +c1x1+c2. (2.135) Applying the boundary conditions for a clamped–free beam from equation (2.131) results inc1=c2 =0. Substituting this result into equation (2.135) and writing the expression for the deflection results in
u3(x1)= f YI33
1
2L x12−1
6x13 . (2.136)
The result of the analysis is that the deflection of a clamped–free beam with an end load is a cubic equation inx1. This contrasts with the result for a constant moment, which was a quadratic expression. The stiffness of the beam at the point of loading is obtained by substitutingx1=Linto equation (2.135) and solving for the force in
terms of the displacement:
f
u3(L) =3YI33
L3 . (2.137)
Now consider the case in which the point load acts at the center of a pinned–pinned beam. In this case the moment induced by the point load is a piecewise continuous function of the form
M2(x1)=
f
2x1=YI33d2u3
d2x1
0≤x1≤ L/2 f
2(L−x1)=YI33d2u3
d2x1
L/2≤x1 ≤L.
(2.138)
Substituting the two expressions into the flexure equation and integrating once yields
2YI33du3
d x1
=
1
2f x12+c1 0≤x1≤ L/2 f
L x1−1
2x12 +c3 L/2≤x1 ≤L. (2.139) Integrating a second time yields
2YI33u3 =
1
6f x13+c1x1+c2 0≤x1≤ L/2 f
1
2L x12−1
6x31 +c3x1+c4 L/2≤x1≤L. (2.140) Notice that the solution requires solving for four constants of integration. Two of the integration constants are obtained from the boundary conditions for a pinned–pinned beam,u3(0)=u3(L)=0. The remaining two constants of integration are obtained by ensuring continuity of the solutions at the location of the point load,x1=L/2.
Applying these four conditions yields the following system of equations:
c2=0
−c3L−c4= f L3
3 (2.141)
c1L
2 +c2−c3L
2 −c4= f L3 12 c1−c3= f L2
4 .
The system of equations is solved to yield c1 = −f L2
8 c2 =0
(2.142) c3 = −3f L2
8 c4 = f L3
24 .
The solutions for the constants of integration are substituted back into equa- tion (2.140). After simplification, the displacement functions over the two regions of the beam are
u3(x1)=
f 2YI33
1 6x13− L
8x21 0≤x1≤L/2
f 2YI33
−1 6x13+L
2x12−3L2
8 x1+ L3
24 L/2≤x1≤ L.
(2.143) An important feature of the solution for a pinned–pinned beam is that the deflection is expressed as apiecewisecontinuous function over the length of the beam. Enforcing the continuity conditions ensures that the solution and its first derivative are continuous at the location of the point load. At the application of the point load the second derivative of the solution is discontinuous and the solution in each of the two regions is described by equation (2.138). This contrasts with the results for the cases studied previously, that of a constant moment and a point load for a clamped–free beam, where the solution for the moment is a continuous function over the domain of the beam. This issue will be important in future chapters when we study techniques for approximating the solution of smart material systems using energy methods.
The stiffness at the centerpoint of the pinned–pinned beam is obtained by substi- tutingx1 =L/2 into equation (2.143) and solving for the ratio of the force to the deflection. The result is
f
u3(L/2) =48YI33
L3 . (2.144)
Comparing this result with equation (2.137), we see that a pinned–pinned beam has a stiffness that is 16 times larger than the stiffness of a cantilever beam with a point load at the free end.
To complete the analysis of bending beams, let us determine the strain energy function. Since we have assumed that each differential element is in a state of uniaxial stress and have neglected any shear stress in the beam, the strain energy per unit volume is equal to the strain energy for an axial bar. Substituting the expression for
the bending strain, equation (2.123), into equation (2.119) yields U˜ = 1
2x32Y d2u3
d2x1
2
. (2.145)
Note that the stored energy per unit volume changes through the thickness of the beam because the strain changes through the thickness. The total stored energy is obtained by integrating equation (2.145) throughout the volume. This integral is separated in the following fashion:
U= L
0
1 2Y
d2u3 d2x1
2
A
x23d x2d x3d x1. (2.146) The area integral is equal to our definition of the second area moment of inertia.
Assuming that both the modulus and the area moment of inertial do not change along the length, we can write the total stored energy as
U= 1 2YI33
L 0
d2u3 d2x1
2
d x1. (2.147)
Equation (2.147) is a general expression that applies to any beam in bending that is consistent with the assumptions of the analysis. The exact expression for the stored energy will be a function of the loading and the boundary conditions. The expression can be solved explicitly with the solutions for the displacement functions obtained in this section.